## Triple Cube

Find all integers between 100 and 1000 such that each digit is a different odd number and the sum of the cubes of the digits is the original three-digit number.
Source: mathcontest.olemiss.edu 2/9/2009

SOLUTION
If you try to construct 3-digit numbers with different odd digits 1, 3, 5, 7, and 9, you can make $5\times 4\times 3=60$ 3-digit numbers. For example, 135, 137, 139, etc. up to 971, 972, 975.

A long way to solve the problem is to go through the list of 60 numbers and calculate the cube of the digits and see if their sum is equal to the original number. For example,
$1^3+3^3+5^3=153\neq 135$
$1^3+3^3+7^3=371\neq 137$
$\cdots$

A faster way is to calculate the sums of the cubes $1^3, 3^3, 5^3, 7^3, \textup{and }9^3$ and see if the sums are equal to the original numbers.

How many possible sums are there? Since we are choosing 3 cubes from a set of 5 cubes and we do not care about order because addition is commutative, we have 10 sums as described below:
$1^3+3^3+5^3=153$
$1^3+3^3+7^3=371$
$1^3+3^3+9^3=757$
$1^3+5^3+7^3=469$
$1^3+5^3+9^3=855$
$1^3+7^3+9^3=1073$
$3^3+5^3+7^3=495$
$3^3+5^3+9^3=881$
$3^3+7^3+9^3=1099$
$5^3+7^3+9^3=1197$

We can see right away that there are two solutions:
$1^3+5^3+3^3=1^3+3^3+5^3=153$
$3^3+7^3+1^3=1^3+3^3+7^3=371$